Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
double eps = 1e-7;
long long INF = 1e9;
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  return a.second < b.second;
}
int main() {
  int n, h;
  cin >> n >> h;
  double s = (h * 1.0) / 2.0;
  double one = s / (double)n;
  vector<double> x(n);
  double prev = 0;
  double s2 = 0;
  for (int i = 0; i < n; i++) {
    double l = prev;
    double r = h;
    for (int j = 0; j < 200; j++) {
      double mid = (l + r) / 2.0;
      double s1 = mid / (double)h;
      double h2 = mid - prev;
      double S = (s1 + s2) / 2.0 * h2;
      if (S > one) {
        r = mid;
      } else {
        l = mid;
      }
    }
    x[i] = l;
    prev = l;
    s2 = l / (double)h;
  }
  for (int i = 0; i < n - 1; i++) {
    cout << fixed << setprecision(15) << x[i] << " ";
  }
}